寻找两个正序数组的中位数
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23
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23
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@ -13,11 +13,32 @@
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<attribute name="test" value="true"/>
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<classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/JavaSE-1.8"/>
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<classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/JavaSE-1.8">
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<attributes>
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<attribute name="maven.pomderived" value="true"/>
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<classpathentry kind="con" path="org.eclipse.m2e.MAVEN2_CLASSPATH_CONTAINER">
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<attributes>
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<attribute name="maven.pomderived" value="true"/>
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</attributes>
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</classpathentry>
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<classpathentry kind="src" path="target/generated-sources/annotations">
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<attributes>
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<attribute name="optional" value="true"/>
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<attribute name="maven.pomderived" value="true"/>
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<attribute name="ignore_optional_problems" value="true"/>
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<attribute name="m2e-apt" value="true"/>
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</classpathentry>
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<classpathentry kind="src" output="target/test-classes" path="target/generated-test-sources/test-annotations">
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<attributes>
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<attribute name="optional" value="true"/>
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<attribute name="maven.pomderived" value="true"/>
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<attribute name="ignore_optional_problems" value="true"/>
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<attribute name="m2e-apt" value="true"/>
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<attribute name="test" value="true"/>
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</attributes>
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</classpathentry>
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<classpathentry kind="output" path="target/classes"/>
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</classpath>
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11
.project
11
.project
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<nature>org.eclipse.jdt.core.javanature</nature>
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<nature>org.eclipse.m2e.core.maven2Nature</nature>
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</natures>
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<filteredResources>
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<filter>
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<id>1605433362927</id>
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<name></name>
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<type>30</type>
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<matcher>
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<id>org.eclipse.core.resources.regexFilterMatcher</id>
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<arguments>node_modules|.git|__CREATED_BY_JAVA_LANGUAGE_SERVER__</arguments>
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</matcher>
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</filter>
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</filteredResources>
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</projectDescription>
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@ -0,0 +1,2 @@
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eclipse.preferences.version=1
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org.eclipse.jdt.apt.aptEnabled=false
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@ -11,5 +11,6 @@ org.eclipse.jdt.core.compiler.problem.enablePreviewFeatures=disabled
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org.eclipse.jdt.core.compiler.problem.enumIdentifier=error
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org.eclipse.jdt.core.compiler.problem.forbiddenReference=warning
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org.eclipse.jdt.core.compiler.problem.reportPreviewFeatures=ignore
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org.eclipse.jdt.core.compiler.processAnnotations=disabled
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org.eclipse.jdt.core.compiler.release=disabled
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org.eclipse.jdt.core.compiler.source=1.8
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@ -2,8 +2,8 @@
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日常刷lintCode算法,提高自己的编码能力.
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- [字符串相关](./src/string)
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- [list相关](./src.list)
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- [简单算法](./src/simple)
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- [九章算法](./src/jiuzhang)
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- [字符串相关](./src/main/java/com/leetcode/string)
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- [list相关](./src/main/java/com/leetcode/list)
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- [简单算法](./src/main/java/com/leetcode/simple)
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- [九章算法](./src/main/java/com/leetcode/jiuizhang)
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@ -1,60 +0,0 @@
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## 题目描述
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给定一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？找出所有满足条件且不重复的三元组。
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注意：答案中不可以包含重复的三元组。
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示例:
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给定数组 nums = [-1, 0, 1, 2, -1, -4]，
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满足要求的三元组集合为：
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```
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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```
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## 解题思路
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## 答案
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```java
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class Solution {
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public List<List<Integer>> threeSum(int[] nums) {
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List<List<Integer>> res = new ArrayList<>();
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Arrays.sort(nums);
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int min=0,mid=1, max=nums.length-1, s= 0;
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while(nums.length >= 3 && min < nums.length && nums[min] <= 0){
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mid = min + 1;
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max = nums.length-1;
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if(min > 0 && nums[min] == nums[min -1]){
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min ++;
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continue;
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}
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while(mid < max){
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s = nums[min] + nums[mid] + nums[max];
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if (s < 0){
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mid ++;
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}else if(s > 0){
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max --;
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}else if (s == 0){
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res.add(Arrays.asList(nums[min],nums[mid], nums[max]));
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while(mid < max && nums[mid] == nums[mid + 1]){
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mid ++;
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}
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while(mid< max && nums[max] == nums[max-1]){
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max --;
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}
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mid ++;
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max --;
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}
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}
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min++;
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}
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return res;
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}
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}
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```
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@ -18,7 +18,7 @@ public class PlusAB {
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return a;
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}
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public static void main(String[] args){
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public static void main(String[] args) {
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PlusAB ab = new PlusAB();
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System.out.println(ab.plus(1,3));
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}
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@ -0,0 +1,70 @@
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package com.leetcode.list;
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import java.text.DecimalFormat;
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/**
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*
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*
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给定两个大小为 m 和 n 的正序（从小到大）数组 nums1 和 nums2。请你找出并返回这两个正序数组的中位数。
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进阶：你能设计一个时间复杂度为 O(log (m+n)) 的算法解决此问题吗？
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示例 1：
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输入：nums1 = [1,3], nums2 = [2]
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输出：2.00000
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解释：合并数组 = [1,2,3] ，中位数 2
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示例 2：
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输入：nums1 = [1,2], nums2 = [3,4]
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输出：2.50000
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解释：合并数组 = [1,2,3,4] ，中位数 (2 + 3) / 2 = 2.5
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示例 3：
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输入：nums1 = [0,0], nums2 = [0,0]
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输出：0.00000
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示例 4：
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输入：nums1 = [], nums2 = [1]
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输出：1.00000
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示例 5：
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输入：nums1 = [2], nums2 = []
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输出：2.00000
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题库位置：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/
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*
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* @author zeekling
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*
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*/
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public class FindMedianSortedArrays {
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public double median(int[] nums1, int[] nums2) {
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int all = nums1.length + nums2.length;
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int mi = 0,ni = 0, curr = 0, pre = 0;
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for ( int i = 0;i < all;i++ ) {
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pre = curr;
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if (mi < nums1.length && ni >= nums2.length ){
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curr = nums1[mi++];
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}else if (mi >= nums1.length && ni < nums2.length) {
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curr = nums2[ni++];
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}else {
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curr = nums1[mi] < nums2[ni] ? nums1[mi++] : nums2[ni++];
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}
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if (i*2 == all ){
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return ((double)curr + (double)pre)/2;
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}
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if ((i+1)*2 > all) {
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return curr;
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}
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}
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return 0;
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}
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public static void main(String[] args) {
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FindMedianSortedArrays median = new FindMedianSortedArrays();
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int[] nums1 = {2}, nums2 = {};
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double res = median.median(nums1, nums2);
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DecimalFormat format = new DecimalFormat("#.00000");
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System.out.println(format.format(res));
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}
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}
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@ -17,9 +17,8 @@ public class PivotIndex {
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for (int i=0; i< len;i++){
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sum += nums[i];
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}
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int idx = -1;
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int leftSum = 0;
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for (int i=0;i<len;i++){
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for (int i = 0;i < len;i ++){
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if ((leftSum * 2 + nums[i]) == sum){
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return i;
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@ -11,7 +11,8 @@ package com.leetcode.num;
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public class SumNums{
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public int sumNums(int n) {
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boolean res = n > 0 && (n += sumNums(n-1)) > 0;
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@SuppressWarnings("unused")
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boolean res = n > 0 && (n += sumNums(n-1)) > 0;
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return n;
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}
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